IMPULSE AND MOMENTUM 259 205. Case II. Rough Contact. When the plane is rough frictional forces come into play and change the tangential component of the momentum. Let F be the tangential force due to friction, N the normal force, and & the coefficient of friction; theu we have X = -mvn, '= / Ndt = -JT Lt = / F dt I fjN dt = JQ JQ f*T' f*T' L/= I Fdt= I vN dt = efj,mvn. JT JT But Lt+ L^ = mo I mvt. Therefore m (vtf vt) = ra^ (1 + e) vn and vtf =Vtn(l+e) vn. Substituting this value of vt' in the expression for tan which is obtained from Fig. 121, we get vn evn Eliminating vt between the last equation and the relation tan a = ~ we obtain vn e tan 0 = tan a - M (1 + e). (XV) DISCUSSION. When AC = 0, equation (XV) reduces to equation (XIV). When M = oo , tan /3= ooor/3 = -; therefore the particle slides along 2i the plane towards the left. When e 0 and tan a > p, tan /3 = oo and ft = ~; therefore the particle slides along the plane towards the right.